Shear Walls Shear Walls - EF

Shear Walls • • • • • •

Types of Shear Walls Shear Wall Stiffness Maximum Reinforcement Requirements Shear Strength Example: simple building Shear walls with openings

Shear Walls

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Shear Walls _____________ (unreinforced) shear wall (1.17.3.2.2): Unreinforced wall _____________ (unreinforced) shear wall (1.17.3.2.3): Unreinforced wall with ith prescriptive i ti reinforcement. i f t Structurally connected floor and roof levels

Within 16 in. of top of wall

40db or 24 in. ≤ 8 in.

≤ 8 iin. Corners and end of walls

≤ 16 iin.

Control joint

≤ 10 ft.

Joint reinforcement at 16 in. o.c. or bond beams at 10 ft. Reinforcement not required at openings smaller than 16 in. in in either vertical or horizontal direction

Reinforcement of at least 0.2 in2 Shear Walls

2

Shear Walls _____________ reinforced shear wall (1.17.3.2.4): Reinforced wall with prescriptive reinforcement of detailed plain shear wall. ______________ reinforced shear wall (1.17.3.2.5): Reinforced wall with prescriptive reinforcement of detailed plain shear wall. Spacing of vertical reinforcement reduced to 48 inches. ___________ reinforced shear wall (1.17.3.2.6): 1. 2. 3. 4. 5.

Maximum spacing of vertical and horizontal reinforcement is min{1/3 length of g of wall,, 48 in. [24 [ in. for masonryy in other than running g bond]}. ]} wall,, 1/3 height Minimum area of vertical reinforcement is 1/3 area of shear reinforcement Shear reinforcement anchored around vertical reinforcing with standard hook Sum of area of vertical and horizontal reinforcement shall be 0.002 times gross cross sectional area of wall cross-sectional Minimum area of reinforcement in either direction shall be 0.0007 times gross cross-sectional area of wall [0.0015 for horizontal reinforcement for masonry in other than running bond].

Shear Walls

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Minimum Reinforcement of Special Shear W ll Walls Reinforcement Ratio 0.0007

0.0010

0.0013

8 in. CMU wall As (in2/ft)

Possibilities

0.064

#4@32 #5@56

12 in. CMU wall As (in2/ft)

Possibilities

0.098

#4@24 #5@32 #6@48

0.092

@ #4@24 #5@32 #6@40

0.140

#4@16 @ #5@24 #6@32

0.119

#4@16 #5@32 #6@40

0.181

#5@16 #6@24

Use specified dimensions, e.g. 7.625 in. for 8 in. CMU walls. Shear Walls

4

Special Shear Walls: Design Minimum strength, strength design (1.17.3.2.6.1.1): Design shear strength, Vn, greater than 1.25 times nominal flexural strength strength, Mn, except Vn need not be greater than 2.5Vu.

Shear Walls

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Seismic Design Category Seismic Design Category

Allowed Shear Walls

A or B C D and higher Shear Wall

Response modification factor: Seismic design force divided by response modification factor, which accounts for ductility and energy absorption. Knoxville, Tennessee Category C for Use Group I and II Category D for Use Group III (essential facilities

R

Ordinary plain

1.5

Detailed plain

2

Ordinary reinforced

2

Intermediate reinforced

3.5

Special reinforced

Shear Walls

5

6

Shear Walls: Stiffness

h

d

h/d < 0.25

0.25 < h/d < 4.0

_____ stiffness predominates

Both shear and bending stiffness are important

h/d >4.0

_______ stiffness predominates

Shear Walls

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Coupled Shear Walls

________ Shear Sh W Wallll

___________ Shear Wall Shear Walls

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Shear Wall Stiffness 3.1.5.1 Deflection calculations shall be based on cracked section properties. Assumed properties shall not exceed half of gross section properties, unless a cracked-section analysis is performed. Cantilever wall

 cant 

kcant 

Et 2   h   h    4   3  L    L  

Fi d wallll (fixed against rotation at top) Fixed

 fixed 

k fixed 

h = height of wall Av = shear area; 5/6A for a rectangle G = shear modulus; given as 0.4E (1.8.2.2.2) t = thickness hi k off wallll L = length of wall

Et 2   h   h   3       L   L  

Real wall is probably between two cases; diaphragm provides some rotational restraint, but not full fixity. Shear Walls

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T- or L- Shaped Shear Walls Section 1.9.4 Wall intersections designed either to: a) ____________________: b) ____________________ Connection that transfers shear: (must be in running bond) a) Fifty percent of masonry units interlock b) Steel connectors at max 4ft. c) Intersecting bond beams at max 4 ft. ft Reinforcing of at least 0.1in 0 1in2 per foot of wall

Metal lath or wire screen to support grout

2-#4’s

1/ in. x 11/ in. x 28in. 4 2 with 2in. long 90 deg bends at each end to form U or Z shape

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Effective Flange Width (1.9.4.2.3) Effective flange width on either side of web shall be smaller of actual flange width, distance to a movement joint, or: • Flange in compression: 6t • Flange in tension: • Unreinforced masonry: 6t • Reinforced masonry: 0.75 times floor-to-floor wall height

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Maximum reinforcing (3.3.3.5) No limits on maximum reinforcing for following case (3.3.3.5.4): Mu  1 and R  1.5 Squat walls, not designed for ductility Vu d v In other cases, can design by either providing boundary elements or limiting reinforcement. Boundary element design (3.3.6.5): More difficult with masonry than concrete Boundary elements not required if:

Pu  0.1 f m A g

geometrically symmetrical sections

Pu  0.05 f m A g

geometrically unsymmetrical sections

AND Mu 1 Vu lw

OR

Vu  3 An f m

Shear Walls

AND

Mu 3 Vu lw 13

Maximum reinforcing (3.3.3.5) Reinforcement limits: Calculated using Maximum stress in steel of fy Axial forces taken from load combination D+0.75L+0.525Q D 0.75L 0.525QE Compression reinforcement, with or without lateral ties, permitted to be included for calculation of maximum flexural tensile reinforcement

Uniformly distributed reinforcement  = 1.5 ordinary walls  = 3 intermediate walls  = 4 special walls

  mu  P  0.64 f m b   d    As y  v  mu  dv      f y  y mu       y   mu   mu  P  0.64 f m b      bd As mu y     d bd   f y  min  mu   mu   y ,  y  Es  d  

Compression steel with area equal to tension steel

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Maximum reinforcing Consider a wall with uniformly distributed steel: y Cm  Cs  Ts  P y mu

Strain fy

0.8f’m

Steel in tension

Stress

Steel in compression

   mu Cm  0.8 f m  0.8 d v b      mu y     y   y   y 1  y   Ts  f y As           2   mu y y y   Portion of steel in tension

Yielded steel

Elastic steel

  mu    mu   y 1  y   Cs  f y As           2  mu  mu y mu      mu    mu  0.5 y    f y As          mu y    mu

As taken as total steel dv is actual depth of masonry Shear Walls

  mu  0.5 y    f y As       mu y  

15

Maximum reinforcing Ts  Cs  Cm  P   y  0.5 y     0.5 y      mu    f y As  mu   f y As  y  Ts  C s  f y As                 mu y mu y mu y         y   mu      0.8 f m  0.8  mu Ts  Cs  f y As  d v b  P  Cm  P        y   mu   mu   y 

  mu  P  0.64bf m    d    As y  v  mu  dv     mu   fy y      mu y   Shear Walls

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Maximum reinforcing Maximum steel on a per foot length of wall basis given for 8 in. wall

0.03

0.06

3 2.5

CMU f'm =2ksi

Clay f'm =2ksi

0.015

1.5

CMU f'm =1.5ksi

0.01 0.005

4

0.04 2

0.02

2

As/(bdv)

Clay f'm =4ksi

3.5 3

0.03

2.5

CMU f'm =2ksi

1

0.02

0.5

0.01

2 15 1.5 1 0.5

CMU f'm =1.5ksi

0

0

0

0.05

0.1

0.15

0

0.2

0

0

[P/(bd)]/f'm

0.05

0.1

0.15

0.2

0.25

0.3

[P/(bd)]/f'm

Shear Walls

5 4.5

Clay f'm =2ksi

As (in /ft)

As/(bdv)

0.025

Clay f'm =4ksi

 s=2 y 0.05

17

2

 s=4 y

As (in /ft)

0.035

Shear Strength (3.3.4.1.2)   0.8

Vn  ___ ____

  M  Vm  4.0  1.75 u  An  Vu d v   A  Vs  0.5 v  f y d v  s  Maximum Vn is:

Mu/Vudv need d nott be b ttaken k >1 1.0 0 Pu = axial load

f m  0.25Pu

Vertical reinforcement shall not be less than onethird horizontal reinforcement; reinforcement shall be uniformly distributed, max spacing of 8 ft (3.3.6.2)

Vn  6 An

f m

( M u / Vu d v )  0.25

Vn  4 An

f m

( M u / Vu d v )  1.0

Interpolate for values of Mu/Vudv between 0 0.25 25 and 1 1.0 0   M 4 Vn   5  2 u  An f m Vu d v  3 Shear Walls 18

Example Given: 10ft. high x 16 ft long 8 in. CMU shear wall; Grade 60 steel, f’m=1500psi; 2-#5 each end; #5 at 48in. (one space will actually be 40 in.); superimposed dead load of 1kip/ft. R Required: i d Maximum M i wind i d lload d based b d on fl flexurall capacity; it shear h steel t l required to achieve this capacity. Solution: Check capacity in flexure using 0.9D+1.0W load combination. Weight of wall: 38 psf(10ft) = 0.38 k/ft (Wall weight from lightweight units, grout at 48 in. o.c.)

Pu=0.9(1k/ft+0.38k/ft)=1.24k/ft 0 9(1k/ft 0 38k/ft) 1 24k/ft

T t l 1.24k/ft(16ft) Total: 1 24k/ft(16ft) = 19 19.88kips 88ki

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Example Solve for stresses, forces and moment in terms of c, depth to neutral axis. T1

T2

T3

T4

C2

c

C1

8 in

52 in 92 in 140 in 188 in

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Example Solve for stresses, forces and moment in terms of c, depth to neutral axis. T1

T2

T3

T4

C2

c

C1

8 in

52 in 92 in 140 in 188 in

• Use trial and error to find c such that Pn = 19.88 kips/0.9 = 22.09 kips • Pn = C1 + C2 – T1 – T2 – T3 – T4 • c = 28.2 inches • Sum moments about middle of wall (8 ft from end) to find Mn • Mn = 1123 kip kip-ft; ft; Mu = 0.9(1123 0 9(1123 kip kip-ft) ft) = 1010 kip kip-ft ft • Wind = Mu/h = (1010 kip-ft)/(10ft) = 101 kips Shear Walls

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Example Calculate net area, An, including grouted cells.

Net area

Net area:

M u / Vu d v   Vn  4 An

Maximum Vn

f m

M u / Vu d v   0.25

M / V d  0.25   Vn   6  2 u u v  An 1.0  0.25  

Vn  6 An

f m

M u / Vu d v   1.0

f m 

Vu  0.8Vn   0.8132.6kips   106.1kips

Maximum Vu

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Example   M  Vm  4.0  1.75 u  An f m  0.25 Pu  Vu d v  

Pu = 0.9(1k/ft) = 0.9k/ft Total: 0.9k/ft(16ft) = 14.4 kips

Vs  Vn  Vm  101k / 0.8  80.7 k  45.6k Use #5 bars in bond beams beams.

A  Vs  0.5 v  f y d v  s 



s

Determine spacing spacing.

0.5 Av f y d v Vs



Use 4 bond beams, spaced at 32 in. o.c. vertically. Actual construction has 15 courses: Bond beam in course 4, 8, 12, and 15 from bottom Shear Walls

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Example: Maximum Reinforcing 3.3.3.5 No limit to reinforcement since Mu/(Vudv) < 1. If we needed to check maximum reinforcing, g, there are at least two ways. y Check on axial load: a) Set steel strain to limit and find c b) Find axial load for this c c) If applied axial load is less than axial load, OK

Steel strain = 1.5y = 0.00310 c = 0.0025/(0.0025+0.00310)*192 0 0025/(0 0025+0 00310)*192 = 85 85.7 7 in in. From spreadsheet, Pu = 181 kips OK

Find steel strain for given axial load: a)) Through Th h ttrial i l and d error ((or solver), l ) find c for given axial load b) Determine steel strain c)) If steel strain g greater than limit,, OK

P = 22 22.08 08 kips (Dead load) c = 28.46 in. s = (192-28.5)/28.5*0.0025 = 0.0143 = 6.93y OK

Note: did not include compression steel which is allowed when checking max reinforcing

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Example: Issues with Strength Design If this were a special shear wall: • Vn ≥ shear corresponding to 1.25Mn (1.17.3.2.6.1.1) • 1.25Mn = 1.25(1123) = 1404k-ft • Vn = Mn/h = 1404k-ft/10ft = 140.4kips • Vn = 140.4k/0.8 = 175.4kips • But, maximum Vn is 132.6 kips (76% of required). Cannot build wall. Need to take out some flexural steel or fully grout to get greater shear capacity.

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Partially Grouted Shear Walls Recent research has shown that TMS Code equations overestimate capacity of partially grouted shear walls. T Two possible ibl modifications: difi ti •

Only use face shell area, 2.5in(192in) = 480in2 (30% reduction)

•

Multiply An by An/Ag, an empirical reduction

Shear Walls

(53% reduction)

29

Example Given: Wall system constructed with 8 in. grouted CMU (Type S mortar). Required: Determine the shear distribution to each wall

112in

Solution: S l ti U gross Use properties (will not make any difference for shear distribution). distribution)

40in

40in

64in

40in

40in

6t=48in

56in

Elevation

a

b

c

Plan Shear Walls

30

Example Pier b

kb 

Et   h   h    4   3  L    L   2



E 7.62in   0.286inE 2   112 in 112 in       3   4 64 in 64 in      

Piers a and c Determine stiffness from basic principles. A Find centroid from outer flange surface

y I

Av  Awebb  Shear Walls

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Example k a ,c 

1 P P    0.157inE 3 3 Ph  Ph  112in  112in   3EI Av G 3E 86000in 4 305in 2 0.4 E 





Portion of shear load = stiffness of pier over the sum of the stiffnesses.

Va 

ka

k

V

0.157inE V  0.26V 0.157inE  0.286inE  0.157inE

Vb = 0.48V

Deflection calculations: k = (0.600in)E = (0.600in)(1350ksi) = 810kip/in Reduce by a factor of 2 to account for cracking: k = 405kip/in

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33

Example Required: Design the system for a horizontal earthquake load of 35 kips, a dead load of 1200 lb/ft length of building, a live load of 1000 lb/ft length of building, and a vertical earthquake force of 0.2 times the dead load. Use a special reinforced shear wall (R>1.5). Solution: Use strength design. Check deflection.



P 35k   0.086in k 405k / in

 0.086in   0.000768 h 112in 112 in



or

Shear Walls

h 1302

34

Example Wall b: Vub = 0.48Vu = 0.48(35k) = 16.8k D = 1.2k/ft(_____________)(1ft/12in) + 0.075ksf(112in)(64in)(1ft2/144in2) = 14.1kips Tributary width

Use load combination 0.9D+E Mu = 16.8k(112in)(1ft/12in) = 156.8k-ft

Pu = 0.9(14.1k) - 0.2(14.1k) = 9.87k

Minimum prescriptive reinforcement for special shear walls: Sum of area of vertical and horizontal reinforcement shall be 0.002 times gross cross-sectional area of wall Minimum area of reinforcement in either direction shall be 0.0007 times gross cross-sectional area of wall Use prescriptive reinforcement of (0.002-0.0007)Ag = 0.0013Ag in vertical direction As = 0.0013(64in)(7.625in) = 0.63in2

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35

Example: Flexural Steel Use load combination 0.9D+E

Mu = 156.8k-ft

Pu = 9.87k

Try y 2-#4 each end;; 1-#4 middle: As=1.00in2 Middle bar is at 28 in. (8 cells, cannot place bar right in center) From spreadsheet: Mn=150.5k-ft NG N t used Note: d solver l tto fifind d value l

Try 1-#6 each end; 1-#6 middle: As=1.32in2 Mn=183.4k-ft

OK

Try 2-#5 each end; 1-#5 middle: As=1.55in2 Mn=214.1k-ft

OK

Use 2-#5 each end; 1-#5 middle: As=1.59in2

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36

Example: Maximum Reinforcing Check maximum reinforcing: Axial force from load combination D+0.75L+0.525QE L 1k/ft(104in)(1ft/12in) = 8.67kip L=1k/ft(104in)(1ft/12in) 8 67kip D+0.75L+0.525QE = 14.1k+0.75(8.67k)+0.525(0) = 20.60kip For maximum reinforcing calculations, compression steel can be included even if nott laterally l t ll supported t d For #5 bars and P = 20.60kip, c = 7.77in. (from solver seek on spreadsheet)

s 

d  kd 60in  7.77in emu  0.0025  0.0168  8.12 y  4 y kd 7.77in

OK Shear Walls

37

Example: Shear M u Vu 112in    1.75  1 Vu d v Vu 64in 

Use Mu/Vudv = 1.0

Find Pu neglecting wall weight Pu = 0.9(10.4k) - 0.2(10.4k) = 7.28k

Vnm  4.0  1.75M u / Vu d v An f m  0.25Pu  2.25(7.625in)(64in) 1500 psi 1k / 1000lb   0.257.28k   44.3kips

Vn  44.3kips p

Ignore any shear steel:

Vn  0.844.3kips   35.5kips

OK

Shear Walls

38

Example: Shear Mn = Mn/ = 214.1k-ft/0.9 = 237.9k-ft V corresponding to Mn: [237.9k-ft/(112in)](12in/ft) = 25.5kips 1.25 times V: 1.25(25.5k) = 31.9kips < Vn=35.5kips

OK

Minimum prescriptive horizontal reinforcement: 0.0007Ag As = 0.0007(64in)(7.625in) = 0.34in2 Maximum spacing is 48 in in. Use 3 bond beams, 40in. o.c. (top spacing is only 32in.)

Wall b: Vertical bars: Horizontal bars:

2-#5 each end; 1-#5 middle #4 in bond beam at 40 in. o.c. (Total 3-#4)

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39

Example Wall a: Vua = 0.26Vu = 0.26(35k) = 9.1k

D = 1.2k/ft(20in+40in)(1ft/12in) + 0.075ksf(112in)(40in+48in)(1ft2/144in2) = 11.1kips

Use load combination 0.9D+E Mu = 9.1k(112in)(1ft/12in) = 84.9k-ft

Pu = 0.9(11.1k) - 0.2(11.1k) = 7.77k

Minimum prescriptive reinforcement: 0.0013Ag As = 0.0013(88in)(7.625in) = 0.87in2

Design issues:

Flange in tension: easy to get steel; problems with max Flange in compression: hard to get steel; max is easy Steel in both flange and web affect max Need to also consider loads in perpendicular direction Need to check shear at interface of flange and web

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40

Example Use load combination 0.9D+E

Mu = 84.9k-ft

Pu = 7.77k

Axial force for max from load combination D+0.75L+0.525QE L=1k/ft(60in)(1ft/12in) = 5.00kip 5 00kip D+0.75L+0.525QE = 11.1k+0.75(5.00k)+0.525(0) = 14.85kip To be consistent with other wall wall, try #5 bars Try 3 in the flange (either distributed as shown, or one in the corner, and two in the jamb Try 2 in the web jamb

Flange in tension: Mn=146.4k-ft

Max reinforcing check: s=0.0101=4.86y

OK

Note: with 2 2-#5 #5 in flange flange, Mn=104.8k-ft, =104 8k ft but might need 3 bars for perpendicular direction

Flange in compression: Mn=124.3k-ft Max reinforcing check: s=0.0417=20.2y Shear Walls

OK 41

Example: Shear M u Vu 112in    2.8  1 Vu d v Vu 40in 

Use Mu/Vudv = 1.0

Vm  2.25(7.625in)(40in) 1500 psi 1k / 1000lb   0.254.2k   27.6kips Ignore any shear steel:

Vn  27.6kips

Vn  0.827.6kips   22.1kips

Mn = Mn/ = 146.4k-ft/0.9 = 162.7k-ft V corresponding to Mn: [162.7k-ft/(112in)](12in/ft) = 17.4kips 1 25 times V: 1 1.25 1.25(17.4k) 25(17 4k) = 21 21.8kips 8kips < Vn=22.1kips =22 1kips

OK

OK

Minimum prescriptive horizontal reinforcement: 0.0007Ag Use 3 bond beams beams, 40in 40in. o o.c. c (top spacing is only 32in 32in.))

Walls a,c:

Vertical bars: Hori ontal bars Horizontal bars:

5-#5 as shown in previous slide #4 in bond beam at 40 in in. o.c. o c (Total 3 3-#4) #4) Shear Walls

42

Example Section 1.9.4.2.4 requires shear to be checked at interface. Check intersection of flange and web in walls a and c. Approximate shear force is tension force in flexural steel.





Vu  As f y  3 0.31in 2 60ksi   55.8kips

Conservatively take Mu/Vudv=1.0, 1.0, and Pu=0. 0.

Vm  4.0  1.75M / Vd v An f m  0.25 Pu  4.0  1.751.0(7.625in)(112in) 1500 ppsi 1k / 1000lb   74.4kips p

Vn  0.874.4kips   59.5kips

OK

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Example: Drag Struts Distributed shear force:

Vu 35k 12in   1.875k / ft L 224in 1 ft

Look at top of wall: 1.88k/ft 9.1k 3 33f 3.33ft

1.88k/ft 88 / 2.8k Drag strut 3 33f 3.33ft

1.88k/ft 88 / 3.4k

16.8k 5.33ft 33f

1.88k/ft 88 / 3.4k

Drag strut 2.8k

3 33f 3.33ft

Shear Walls

1.88k/ft 88 / 9.1k 3 33f 3.33ft

44

Shear Walls: Building Layout 1. All elements either need to be isolated, or will participate in carrying the load 2. Elements that participate in carrying the load need to be properly detailed for seismic requirements 3. Most shear walls will have openings 4. Can design g only yap portion to carry y shear load, but need to detail rest of structure

Shear Walls

45

Shear Walls with Openings Shear walls with openings will be composed of solid wall portions, and portions with piers between openings. Piers: Pi • Requirements only in strength design • Length ≥ 3 x thickness; width ≤ 6 x thickness • Height < 5 x length • Factored axial compression ≤ 0.3Anf’m (3.3.4.3.1) • One bar in each end cell (3.3.4.3.2) • Minimum reinforcement is 0 0.0007bd 0007bd (3.3.4.3.2) (3 3 4 3 2)

Shear Walls

46

f 

h 3  2kt  kt kb  2kb  3  12h   1    6 EI  kt  2kt kb  kb  5 EA

 kt 1  kb    M t  Vh   k 2 k k k t b b   t  kb 1  kt    M b  Vh  k t  2k t kb  k b  As the top spandrel decreases in height, the top approaches a fixed condition against rotation.

kt 

kb 

h ht h hb

Top spandrel

h

Divide wall into piers. Find flexibility of each pier Stiffness is reciprocal of flexibility Di t ib t load Distribute l d according di tto stiffness tiff

hb

1. 2. 3. 4 4.

ht

Shear Walls with Openings

Bottom spandrel

If hb = 0 h 3  2  kt  12h   1    f  6 EI  2kt  1  5 EA  kt   M t  Vh 2 k  1  t 

 1  kt   M b  Vh 2 1 k   t 

Qamaruddin, M., Al-Oraimi, S., and Hago, A.W. (1996). “Mathematical model for lateral stiffness of shear walls with openings.” Proceedings, Seventh North American Masonry Conference, 605-617. Shear Walls 47

Example - Shear Walls with Openings Required: A. Stiffness of wall B. Forces in each pier under 10 kip horizontal load

4ft 4

4fft

Given:

B

C

8ft

A

2ft

4ft

2ft

6ft

2ft

Shear Walls

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Example - Shear Walls with Openings Sample calculations: Pier B h = 4ft I

f 

ht = 4ft

hb = 8ft kt = h/ht = 4ft/4ft = .0

tL3  12

kb = h/hb = 4ft/8ft = 0.5

A  tL  2 ft t

h 3  2kt  kt kb  2kb  3  12h   1   6 EI  kt  2kt kb  kb  5 EA

k

1 1   0.0210 Et f 47.6 / Et

Shear Walls

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Example - Shear Walls with Openings Pier h (ft) ht (ft) hb (ft) kt

kb

f

I (ft3) A (ft)

s

f

k

A

12

4

0

3

inf 0.667t

2t

308.4 18 326.4/Et 0.0031Et

B

4

4

8

1 0.5 0.667t

2t

41.6

6

47.6/Et 0.0210Et

C

4

4

8

1 0.5 0.667t

2t

41.6

6

47.6/Et 0.0210Et

f is flexural deformation; s is shear deformation Total stiffness is 0.0451Et Average stiffness from finite element analysis 0.0440Et Solid wall stiffness 0 0.1428Et 1428Et (free at top); 0 0.25Et 25Et (fixed at top)

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Example - Shear Walls with Openings Sample calculations: Pier B

Vb 

kb

k

V

0.0210 Et 10k  4.66kips 0.0451Et

 kt 1  kb   M t  Vh   k  k k  k 2 t b b   t  kb 1  kt     M b  Vh  kt  2kt k b  k b 

Pier

V (kips)

Mt (k-ft)

Mb (k-ft)

A

0.68

3.50

4.66

B

4.66

11.19

7.46

C

4.66

11.19

7.46

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Example - Shear Walls with Openings To find axial force in piers, look at FBD of top 4ft of wall.

Moment direction on piers

4ft

10k

0.68k 3.50k-ft 2ft

4.66k 11.19k-ft

4ft

2ft

6ft

4.66k 11.19k-ft 2ft

This 65.9k-ft moment is carried by axial forces in the piers. Axial forces are d t determined i d ffrom M Mc/I, /I where h each h pier i iis considered id d a concentrated t t d area. Find centroid from left end.

y

y A A i

i



i

Shear Walls

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Example - Shear Walls with Openings The axial force in pier A is: 4.45k FBD’s FBD s of piers:

3.50k-ft 0.68k

0.45k

11.19k-ft 4.66k

B A

4.66k 0.45k 7.46k-ft

4.90k

11.19k-ft 4.66k

C 4.66k 4.90k 7.46k-ft

0.68k 4.45k 4.66k-ft Shear Walls

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Example - Shear Walls with Openings FBD’s of entire wall to obtain forces in bottom right spandrel: 10k

F

x

 10k  0.68k  Fx  0 Fx  9.32k

F

y

 4.45k  Fy F 0 Fy  4.45k

A

B

C

y

z

 10k (16 ft )  (4.66k  ft )

 4.45k (1 ft )  4.45k (11 ft )  Mz  0 Mz  110.8k  ft

x

Z

0.68k 4 66k ft 4.45k 4.66k-ft

M

Fy=4.45k

Fx=9.32k M 110 8k ft Mz=110.8k-ft Shear Walls

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